1. Prof. David R. Jackson ECE Dept. Fall 2014 Notes 7 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1

2. Coulomb’s Law Experimental law: c = speed of light = 2.99792458  10 8 [ m/s ] (defined) Charles-Augustin de Coulomb 2 Here is how we can calculate  0 accurately: (from ECE 3317) x y z r q1q1 q2q2

3. Coulomb’s Law (cont.) Hence: But E 1 = E due to q 1 Note: There is no self-force on charge 2 due to its own electric field. 3 r = location of observer x y z r q1q1 q2q2 r 1 = (0, 0, 0) r 2 = r

4. Generalization ( q 1 not at the origin): Coulomb’s Law (cont.) r 1 = (x 1, y 1, z 1 ) r 2 = (x 2, y 2, z 2 ) q 2 (x 2, y 2, z 2 ) x y z R r2r2 r1r1 q 1 (x 1, y 1, z 1 ) 4

5. Example q 1 = 0.7 [ mC ] located at (3,5,7) [ m ] q 2 = 4.9 [  C ] located at (1,2,1) [ m ] E 1 ( r 2 ) = electric field due to charge q 1, evaluated at point r 2 Find: F 1, F 2 For F 2 : F 1 = force on charge q 1 F 2 = force on charge q 2 R q2q2 q1q1 5

6. Example (cont.) q 1 = 0.7 [ mC ] located at (3,5,7) [ m ] q 2 = 4.9 [  C ] located at (1,2,1) [ m ] 6 R q2q2 q1q1

7. Example (cont.) (Newton’s Law) 7

8. General Case: Multiple Charges q 1 : r 1 = (x 1, y 1, z 1 ) R 1 = r - r 1 q 2 : r 2 = (x 2, y 2, z 2 )... q N : r N = (x N, y N, z N ) R 2 = r - r 2... R N = r - r N E=E 1 +E 2 +…+E N ( superposition ) x y z R1R1 q2q2 q1q1 R2R2 RNRN R3R3 q3q3 qNqN r = (x, y, z) 8

9. Field from Volume Charge 9 r = (x, y, z) x y z R dV´

10. Field from Volume Charge (cont.) 10 x y z r = (x, y, z) R dV´

11. Field from Surface Charge 11 r = (x, y, z) x y z R dS dS´

12. Field from Line Charge 12 x y z r = (x, y, z) R dl

13. Example q 1 = +20 [nC] located at (1,0,0) [m] q 2 = -20 [nC] located at (0,1,0) [ m ] Find E (0,0,1) R 1 = (0,0,1) - (1,0,0) R 2 = (0,0,1) - (0,1,0) r = (0,0,1) x y z R1R1 q 2 = -20 [nC] R2R2 q 1 = +20 [nC] Solution: 13

14. Example (cont.) 14

15. Example Semi-infinite uniform line charge 15 x y z r = (0, 0, h) R  l =  l0 [C/m] Find E ( 0, 0, h )

16. Example (cont.) Note: The upper limit must be greater than the lower limit to keep dl positive. 16

17. Example Find E ( 0, 0, z ) 17  l =  l0 [C/m] ( uniform ) x y z a R r = (0, 0, z) ( rectangular coordinates ) Note: The upper limit must be greater than the lower limit to keep dl positive. so x a y

18. Example (cont.) Note: Hence y x z a R r = (0, 0, z) 18 x y

19. Example (cont.) We can also get these results geometrically, by simply looking at the picture. 19 y x z a R r = (0, 0, z)

20. Example (cont.) 20 Reminder: The upper limit must be greater than the lower limit to keep dl positive.

21. Example (cont.) Also, Hence 21 a x y

22. Example (cont.) Limiting case: a  0 (while the total charge remains constant) (point-charge result) (total charge on ring) Note: If we wish Q to remain fixed, than  l0 must increase as a gets smaller. 22 where